3.98.61 \(\int \frac {72 x^2 \log (x)+e^2 (400 x+80 x^2) \log ^2(x)+(-360 x-72 x^2+(720 x+144 x^2) \log (x)) \log (5+x)}{(25+5 x) \log ^2(x)} \, dx\)

Optimal. Leaf size=25 \[ 8 \left (e^2 x^2+\frac {9 x^2 \log (5+x)}{5 \log (x)}\right ) \]

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Rubi [F]  time = 0.52, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {72 x^2 \log (x)+e^2 \left (400 x+80 x^2\right ) \log ^2(x)+\left (-360 x-72 x^2+\left (720 x+144 x^2\right ) \log (x)\right ) \log (5+x)}{(25+5 x) \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(72*x^2*Log[x] + E^2*(400*x + 80*x^2)*Log[x]^2 + (-360*x - 72*x^2 + (720*x + 144*x^2)*Log[x])*Log[5 + x])/
((25 + 5*x)*Log[x]^2),x]

[Out]

8*E^2*x^2 + (72*Defer[Int][x^2/((5 + x)*Log[x]), x])/5 - (72*Defer[Int][(x*Log[5 + x])/Log[x]^2, x])/5 + (144*
Defer[Int][(x*Log[5 + x])/Log[x], x])/5

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {8}{5} x \left (10 e^2-\frac {9 \log (5+x)}{\log ^2(x)}+\frac {9 (x+2 (5+x) \log (5+x))}{(5+x) \log (x)}\right ) \, dx\\ &=\frac {8}{5} \int x \left (10 e^2-\frac {9 \log (5+x)}{\log ^2(x)}+\frac {9 (x+2 (5+x) \log (5+x))}{(5+x) \log (x)}\right ) \, dx\\ &=\frac {8}{5} \int \left (\frac {x \left (9 x+50 e^2 \log (x)+10 e^2 x \log (x)\right )}{(5+x) \log (x)}+\frac {9 x (-1+2 \log (x)) \log (5+x)}{\log ^2(x)}\right ) \, dx\\ &=\frac {8}{5} \int \frac {x \left (9 x+50 e^2 \log (x)+10 e^2 x \log (x)\right )}{(5+x) \log (x)} \, dx+\frac {72}{5} \int \frac {x (-1+2 \log (x)) \log (5+x)}{\log ^2(x)} \, dx\\ &=\frac {8}{5} \int x \left (10 e^2+\frac {9 x}{(5+x) \log (x)}\right ) \, dx+\frac {72}{5} \int \left (-\frac {x \log (5+x)}{\log ^2(x)}+\frac {2 x \log (5+x)}{\log (x)}\right ) \, dx\\ &=\frac {8}{5} \int \left (10 e^2 x+\frac {9 x^2}{(5+x) \log (x)}\right ) \, dx-\frac {72}{5} \int \frac {x \log (5+x)}{\log ^2(x)} \, dx+\frac {144}{5} \int \frac {x \log (5+x)}{\log (x)} \, dx\\ &=8 e^2 x^2+\frac {72}{5} \int \frac {x^2}{(5+x) \log (x)} \, dx-\frac {72}{5} \int \frac {x \log (5+x)}{\log ^2(x)} \, dx+\frac {144}{5} \int \frac {x \log (5+x)}{\log (x)} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.10, size = 26, normalized size = 1.04 \begin {gather*} \frac {8}{5} \left (5 e^2 x^2+\frac {9 x^2 \log (5+x)}{\log (x)}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(72*x^2*Log[x] + E^2*(400*x + 80*x^2)*Log[x]^2 + (-360*x - 72*x^2 + (720*x + 144*x^2)*Log[x])*Log[5
+ x])/((25 + 5*x)*Log[x]^2),x]

[Out]

(8*(5*E^2*x^2 + (9*x^2*Log[5 + x])/Log[x]))/5

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fricas [A]  time = 0.64, size = 25, normalized size = 1.00 \begin {gather*} \frac {8 \, {\left (5 \, x^{2} e^{2} \log \relax (x) + 9 \, x^{2} \log \left (x + 5\right )\right )}}{5 \, \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((144*x^2+720*x)*log(x)-72*x^2-360*x)*log(5+x)+(80*x^2+400*x)*exp(2)*log(x)^2+72*x^2*log(x))/(25+5*
x)/log(x)^2,x, algorithm="fricas")

[Out]

8/5*(5*x^2*e^2*log(x) + 9*x^2*log(x + 5))/log(x)

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giac [A]  time = 0.16, size = 25, normalized size = 1.00 \begin {gather*} \frac {8 \, {\left (5 \, x^{2} e^{2} \log \relax (x) + 9 \, x^{2} \log \left (x + 5\right )\right )}}{5 \, \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((144*x^2+720*x)*log(x)-72*x^2-360*x)*log(5+x)+(80*x^2+400*x)*exp(2)*log(x)^2+72*x^2*log(x))/(25+5*
x)/log(x)^2,x, algorithm="giac")

[Out]

8/5*(5*x^2*e^2*log(x) + 9*x^2*log(x + 5))/log(x)

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maple [A]  time = 0.09, size = 22, normalized size = 0.88

method	result	size
risch	\(8 x^{2} {\mathrm e}^{2}+\frac {72 x^{2} \ln \left (5+x \right )}{5 \ln \relax (x )}\)	\(22\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((144*x^2+720*x)*ln(x)-72*x^2-360*x)*ln(5+x)+(80*x^2+400*x)*exp(2)*ln(x)^2+72*x^2*ln(x))/(25+5*x)/ln(x)^2
,x,method=_RETURNVERBOSE)

[Out]

8*x^2*exp(2)+72/5*x^2/ln(x)*ln(5+x)

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maxima [A]  time = 0.46, size = 25, normalized size = 1.00 \begin {gather*} \frac {8 \, {\left (5 \, x^{2} e^{2} \log \relax (x) + 9 \, x^{2} \log \left (x + 5\right )\right )}}{5 \, \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((144*x^2+720*x)*log(x)-72*x^2-360*x)*log(5+x)+(80*x^2+400*x)*exp(2)*log(x)^2+72*x^2*log(x))/(25+5*
x)/log(x)^2,x, algorithm="maxima")

[Out]

8/5*(5*x^2*e^2*log(x) + 9*x^2*log(x + 5))/log(x)

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mupad [B]  time = 6.16, size = 21, normalized size = 0.84 \begin {gather*} 8\,x^2\,{\mathrm {e}}^2+\frac {72\,x^2\,\ln \left (x+5\right )}{5\,\ln \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((72*x^2*log(x) - log(x + 5)*(360*x - log(x)*(720*x + 144*x^2) + 72*x^2) + exp(2)*log(x)^2*(400*x + 80*x^2)
)/(log(x)^2*(5*x + 25)),x)

[Out]

8*x^2*exp(2) + (72*x^2*log(x + 5))/(5*log(x))

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sympy [A]  time = 0.42, size = 22, normalized size = 0.88 \begin {gather*} 8 x^{2} e^{2} + \frac {72 x^{2} \log {\left (x + 5 \right )}}{5 \log {\relax (x )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((144*x**2+720*x)*ln(x)-72*x**2-360*x)*ln(5+x)+(80*x**2+400*x)*exp(2)*ln(x)**2+72*x**2*ln(x))/(25+5
*x)/ln(x)**2,x)

[Out]

8*x**2*exp(2) + 72*x**2*log(x + 5)/(5*log(x))

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